Let the eccentricity e of a hyperbola satisfy the equation 6e^2 - 11e + 3 = 0. If the foci of the hyperbola are (3, 5) and (3, -4), then the length of its latus rectum is :

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Correct answer: C. 15 2

Given the equation for eccentricity: 6e^2 - 11e + 3 = 0 6e^2 - 9e - 2e + 3 = 0 (3e - 1)(2e - 3) = 0 e = 1 3 or e = 3 2 For a hyperbola, e > 1, so e = 3 2 . The distance between the foci (3, 5) and (3, -4) is 2ae. 2ae = (3-3)^2 + (5 - (-4))^2 = 9 2a ( 3 2 ) = 9 3a = 9 a = 3 Using the relation b^2 = a^2(e^2 - 1): b^2 = 9 ( 9 4 - 1 ) = 9 ( 5 4 ) = 45 4 The length of the latus rectum is 2b^2 a : 2 ( 45 4 ) 3 = 45 6 = 15 2 Answer: 15 2

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